32. THEOREM. If two angles of a triangle are unequal, the sides opposite them are unequal, and the greater side lies opposite the greater angle. Let ABC represent the triangle, the angles of which at A and B are unequal. If at the middle point (M) of the side AB a perpendicular be erected, it could not pass through C; for if it did, the triangle would be isosceles. Not passing through C, it must intersect the lines forming the other sides of the triangle at separate points. nearer to M. FIG. 39. B Let D be the intersection that is the Draw the auxiliary line DB. It will separate the angle to which it is drawn into two parts, one of which equals Exercises.-1. Prove the same by using AE as an auxiliary line. 2. Show that the hypothenuse of a right triangle is greater than either of the other sides. 3. Establish the theorem of this section, with an obtuse-angled triangle. 4. Establish the converse of the theorem. 5. Show that if two right triangles have one of the sides adjacent to the right angle and the hypothenuse mutually equal, they are congruent. 33. THEOREM. Any point of an angle bisector is equally distant from the lines forming the angle. If AP represent the angle bisector, and if from any point (P) perpendiculars be let fall to the lines forming the angle, the APAF and PAB will be congruent, having two angles and the included side of one equal to two angles and the included side of the other. Hence PF PB. Q. E. D. THEOREM. The distances of any point, not on the angle bisector, from the angle lines will not be equal. H A K FIG. 41. Let Q represent any point not on the angle bisector. Let QD and QK be the perpendiculars. One of them intersects the angle bisector. Let I be the point of such intersection. Draw HC to AK, and join QC. QD = QH + HD = QH + HC > QC > QK. Exercise. - Show that the bisector of the supplement of KAD will be perpendicular to AI, and that every point in it will be equidistant from the angle lines. NOTE. If perpendiculars on one side of a line are positive (+), those on the other side are negative (−). The locus of a point, the ratio of the distances of which from two lines is (+1), will be the angle bisector If the ratio of the distances is (− 1), the locus will be the bisector of the supplement of the first angle and will be perpendicular to the bisector of the first angle. 34. PROBLEM. To construct a bisector of an angle. If AP were the angle bisector required, and if we should draw the perpendiculars PB and PF, they would be equal to each other, and the segments AB and AF would be equal. If the auxiliary line BF were drawn, AP would be a perpendicular bisector to it. This determination of relations that would exist if the angle bisector were drawn suggests the construction. FIG. 43. Lay off AB equal to AF, and find some point (K), other than (4), which is equally distant from B and F. Draw AK; it will be the angle bisector; since it is the perpendicular bisector of BF. NOTE. We might have followed more closely the relations at first developed, by erecting at B and F perpendiculars to AB and AF, and have joined their intersection with A. 35. If through the point P a perpendicular to the line AB be drawn, and then be rotated positively (i.e. to the left), the point in which it intersects AB will move from Fin the direction of B. When it shall have rotated 90°, be parallel to AB and will not intersect it. We mean the same thing when we say that the point of intersection has passed to infinity. the line through P will The instant the angle exceeds 90°, the rotating line again intersects the line AB, in the direction FA. The distance from F will continually diminish until 180° of rotation has taken place, when the point of intersection will have reached F. The next 180° the point of intersection will travel over the same route as before. QUADRILATERALS AND QUADRANGLES. 36. A plane figure formed by four straight lines which enclose an area forms a quadrilateral. If each line inter D B FIG. 45.-QUADRILATERAL. sects each of the others as indicated in the above figure, the figure is called a complete quadrilateral. The portion ABCD is called a quadrangle, the vertices. of which are at A, B, C, and D, and the sides of which are AB, BC, CD, and DA. # A D FIG. 47. PARALLELOGRAM. If two sides are parallel, the quadrangle is called a trapezoid. |